A lot of what you have written is nonsense (i.e. constructing a homeomorphism from $M \oplus M$ to $S^1 \times \Bbb R$ is clearly impossible (the dimensions don't match) and such a homeomorphism would not tell you the bundle structure). To show a vector bundle is a trivial you have to construct linearly independent sections spanning the fiber at every point. This is easy to do for $M\oplus M$. Really make sure you understand the definitions involved before trying to answer such a question(i.e. vector bundle).
– PVAL-inactiveNov 6 '14 at 15:45

1

@PVAL Thank you for pointing out the mistake, and giving the useful advice.
– gaoxingeNov 7 '14 at 11:15

1 Answer
1

The geometric point is, if $E \to [0, 2\pi]$ is the trivial plane bundle over an interval, and if $U_{1}$ and $U_{2}$ are the non-vanishing sections
$$
U_{1}(t) = (\cos t/2, \sin t/2),\qquad
U_{2}(t) = (-\sin t/2, \cos t/2),
$$
then each $U_{i}$ spans a trivial line subbundle $L_{i}$ of $E$ whose fibre "rotates half a turn between $0$ and $2\pi$", and $L_{1} \oplus L_{2} = E$ as vector bundles over $[0, 2\pi]$. Now map $E$ to the trivial plane bundle over the circle by identifying the fibres over $0$ and $2\pi$, and observe that each $L_{i}$ descends to a non-trivial line bundle over the circle, i.e., to the line bundle $M$ whose total space is a Möbius strip. (The $U_{i}$ themselves do not descend to continuous sections of $M$, of course.)